What is the solution for y^{''}+2y^'+(3/4)y=3e^x+(9/2)x ?

The solution for y^{''}+2y^'+(3/4)y=3e^x+(9/2)x is y=c_{1}e^{-x/2}+c_{2}e^{-(3x)/2}+4/5 e^x+6x-16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+(3/4)y=3e^x+(9/2)x

y^{^{''}} + 2 y^{^{'}} + (\frac{3}{4}) y = 3 e^{x} + (\frac{9}{2}) x

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{3 x}{2}} + \frac{4}{5} e^{x} + 6 x - 16

Solution steps

y^{''} (x) + 2 y^{'} (x) + (\frac{3}{4}) y = 3 e^{x} + (\frac{9}{2}) x

Solve linear ODE:

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{3 x}{2}} + \frac{4}{5} e^{x} + 6 x - 16

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{3 x}{2}} + \frac{4}{5} e^{x} + 6 x - 16

y^{^{''}} + 4 y = cos (x)

(d^{4} + d^{3} + d^{2}) y = 5 x^{2} + cos (x)

(d^{4} + d^{3} + d^{2}) y = 5 x^{2}

(d^{3} - 3 d^{2} + 3 d - 1) y = sin (x) + x^{3}

(d^{3} + d) y = cos (x)

What is the solution for y^{''}+2y^'+(3/4)y=3e^x+(9/2)x ?
The solution for y^{''}+2y^'+(3/4)y=3e^x+(9/2)x is y=c_{1}e^{-x/2}+c_{2}e^{-(3x)/2}+4/5 e^x+6x-16